Question

The face of a dam is shaped like an isosceles trapezoid with a lower base of 44 meters, an upper base of 108 meters, and a height of 114 meters. Find the force on the face of the dam when the water level is 35 meters below the top of the dam.

Answer 1

`F=4047408940.35\ N`

Step-by-step explanation:

Given:

length of upper base of trapezoid,
`u=108\ m`

length of lower base of trapezoid,
`l=44\ m`

height of the trapezoid,
`y=114\ m`

water level from the top of the dam,
`h'=35\ m`

The depth of water from the base of the dam wall:

`d=y-h'`

`d=114-35`

`d=79\ m`

Now the pressure on the dam wall due to water upto the given level:

`P=\rho.g.d`

`P=1000* 9.8* 79`

`P=774200\ Pa`

Now refer the schematic for the area on which the water pressure acts.

Now we have the area as:

`A=(1)/(2)\time (sum\ of\ parallel\ sides)* height`

`A=0.5(44+ 44.3509+44)* 79`

`A=5227.8596\ m^2`

Now the force on the dam wall can be given as:

`F=P.A`

`F=774200* 5227.8596`

`F=4047408940.35\ N`